Question

$\int {\cot }^{5}xdx$

Answer 1

∫ cot⁵ x dx
= ∫ cot⁴ x . cot x dx
= ∫ (cosec² x – 1)² cot x dx
= ∫ (cosec⁴ x – 2 cosec² x + 1) cot x dx
= ∫ cosec⁴ x . cot x dx – 2 ​∫ cot x . cosec² x dx + ​∫ cot x dx
= ∫ cosec² x . cosec² x . cot x . dx – 2 ​∫ cot x cosec² x dx + ∫​ cot x dx
=∫ (1 + cot ² x) . cot x . cosec² x dx – 2 ​∫ cot x cosec² x dx + ​∫ cot x dx
= ∫ (cot x + cot³ x) cosec² x dx – 2 ​∫ cot x cosec² x dx + ​∫ cot x dx
Now, let I₁= ∫ (cot x + cot³ x) cosec² x dx – 2 ​∫ cot x cosec² x dx
And I₂= ∫ cot x dx
First we integrate I₁
I₁= ∫ (cot x + cot³ x) cosec² x dx – 2 ​∫ cot x cosec² x dx
Let cot x = t
⇒ – cosec² x dx = dt
⇒ cosec² x dx = – dt

I₁= ∫ (t + t³) (– dt) – 2​∫ t (–dt)
= –∫(t + t³) + 2​∫t dt
$$\begin{gathered} =\left[-\frac{t^2}{2}-\frac{t^4}{4}\right]+2.\frac{t^2}{2}+C_1 \\ =\frac{t^2}{2}-\frac{t^4}{4}+{\mathrm{C}}_1 \\ =\frac{{\cot }^{2}x}{2}-\frac{{\cot }^{4}x}{4}+C_1 \end{gathered}$$
Now we integrate I₂
I₂= ∫ cot x dx
= $\log \left|\sin x\right|+C_2$
Now, ∫ cot⁵ x dx=I₁ + I₂
= $-\frac{1}{4}{\cot }^{4}x+\frac{1}{2}{\cot }^{2}x+\log \left|\sin x\right|+C_1+C_2$
= $-\frac{1}{4}{\cot }^{4}x+\frac{1}{2}{\cot }^{2}x+\log \left|\sin x\right|+C\left[\therefore C=C_1+C_2\right]$